On a two-lane road, car A is travelling with a speed of 36 km h¯¹. Two cars B and C approach car A in opposite directions with a speed of 54 km h¯¹. each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?
Given the speed of car A and C is 36 km / h and 54 km / h respectively.
The relative speed of car A with respect to car C is,
v AC = 36 km / h + 54 km / h = 90 km / h = 90 km / h × 5 18 = 25 m / s
The relative speed of car B with respect to car A is,
v BA = 54 km / h − 36 km / h = 18 km / h = 18 km / h × 5 18 = 5 m / s
The time taken by the car C to cover the distance of 1 km is,
t = 1000 m v AC
Substitute the required value in the above expression.
t = 1000 m 25 m / s = 40 s
Let s be the distance AC and a be the acceleration of car B.
The second equation of motion is,
s = v BA t + 1 2 a t 2
Substitute the required values in above expression.
1000 m = 5 m / s × 40 s + 1 2 a ( 40 s ) 2 a = 1 m / s 2
Thus, the required minimum acceleration of car B to avoid an accident is 1 ms-2.
Given the speed of car A and C is
The relative speed of car A with respect to car C is,
The relative speed of car B with respect to car A is,
The time taken by the car C to cover the distance of
Substitute the required value in the above expression.
Let
The second equation of motion is,
Substitute the required values in above expression.
Thus, the required minimum acceleration of car B to avoid an accident is 1 ms-2.
